TITLE: Showing that $(\mathbb{Z}/p^{a}\mathbb{Z})^*$ is a cyclic group
QUESTION [6 upvotes]: I'm supposed to show that if $p$ is an odd prime (so its not $2$) that the multiplicative group $(\mathbb{Z}/p^{a}\mathbb{Z})^*$ is cyclic. It seems very hard to do for any $a$, so I'm thinking about using induction on a to make things easy. That being said, I don't really like group theory. How do I show the base case is cyclic and proceed?
Thanks everyone!

REPLY [2 votes]: The post marked answer shows the base case and sets up the induction. Show $(Z/p^2Z)^\times$ is cyclic by showing that if $g$ generates $(Z/pZ)^\times$, then $g$ or $g+p$ generates $(Z/p^2Z)^\times)$.